Are there any ways of solving polynomials with no rational zeros? I attempted to solve the equation $x^5-x+41=0$ using the rational zero theorem. I soon found out that the equation has no rational zeros. Many sources say when presented with a polynomial with no rational zeros, use the Newton approximation method. I didn't want an approximation of the polynomial. Is there a way to solve $x^5-x+41=0$ without approximating? I am aware of the Abel-Ruffini theorem which states that there is no solution in radicals to general polynomial equations of degree five or higher with arbitrary coefficients. Is there possibly a polar or trigonometric solution to this equation?
 A: There is a solution using generalized hypergeometric series.
Writing $x = -41^{1/5} t$, the equation becomes
$$ t^5 - s t - 1 = 0 \tag 1$$
where $s = 41^{-4/5}$.  There is a series for the root of (1) near $t=1$ in powers of $s$:
$$ \eqalign{t &= \sum _{k=0}^{\infty }{\frac {{3125}^{-k}{16}^{k}\Gamma \left( 2\,k-1/
10 \right) \Gamma \left( 2\,k+2/5 \right) \Gamma \left( 4/5 \right) 
\Gamma \left( 3/5 \right) {s}^{5\,k}}{\Gamma \left( -1/10 \right) 
\Gamma \left( 4/5+k \right) \Gamma \left( 3/5+k \right) \Gamma \left( 
2/5+k \right) k!}}\cr &+1/5\,{\frac {{3125}^{-k}{16}^{k}\Gamma \left( 2\,k+
4/5 \right) \Gamma \left( 2\,k+3/10 \right) \Gamma \left( 6/5 \right) 
\Gamma \left( 3/5 \right) {s}^{5\,k+1}}{\Gamma \left( 3/10 \right) 
\Gamma \left( k+6/5 \right) \Gamma \left( 4/5+k \right) \Gamma \left( 
3/5+k \right) k!}}\cr &-1/25\,{\frac {{3125}^{-k}{16}^{k}\Gamma \left( 2\,k
+6/5 \right) \Gamma \left( 7/5 \right) \Gamma \left( 4/5 \right) {s}^{
5\,k+2}}{\Gamma \left( {\frac{7}{10}} \right) \Gamma \left( k+7/5
 \right) \Gamma \left( k+6/5 \right) \Gamma \left( 4/5+k \right) k!}
\Gamma \left( 2\,k+{\frac{7}{10}} \right) }\cr &+{\frac {{3125}^{-k}{16}^{k
}\Gamma \left( 2\,k+8/5 \right) \Gamma \left( 7/5 \right) \Gamma
 \left( 6/5 \right) {s}^{5\,k+3}}{125\,\Gamma \left( {\frac{11}{10}}
 \right) \Gamma \left( k+8/5 \right) \Gamma \left( k+7/5 \right) 
\Gamma \left( k+6/5 \right) k!}\Gamma \left( 2\,k+{\frac{11}{10}}
 \right) }}
$$
